B Observer Moving at c: Does it Contradict Limits?

[jk22]

Does not the movement at c simplifies out depending on how it is approached like :

$$x'=\frac{x-vt}{\sqrt{1-v^2/c^2}}$$

If x=ct, then this gives :

$$x'=c\sqrt{\frac{1-v/c}{1+v/c}}t$$

Then the limit $v\rightarrow c$ exists and implies $x'=0$.

Does this contradict the non existence of observer moving at speed c ?

 

[PeterDonis]

If x=ct, then this gives

Nothing well-defined, because you can't cancel out factors that are zero from numerator and denominator.

Does this contradict the non existence of observer moving at speed c ?

No. It means you are trying to do mathematical operations that aren't allowed. See above.

 

[jk22]

You mean you cannot simplify before taking the limit v->c ? The reverse order if you want.

 

[PAllen]

The expressions before taking the limit are well defined. If you also write the one for t’, you find that x’ = ct’ follows, as expected. However, if you take the limit of both as v goes to c, you get the absurd result that the whole light path is reduced to the single event (0,0). This verifies that a frame moving at c cannot logically exist.

 

[Orodruin]

Does not the movement at c simplifies out depending on how it is approached like :

$$x'=\frac{x-vt}{\sqrt{1-v^2/c^2}}$$

If x=ct, then this gives :

$$x'=c\sqrt{\frac{1-v/c}{1+v/c}}t$$

Then the limit $v\rightarrow c$ exists and implies $x'=0$.

Does this contradict the non existence of observer moving at speed c ?

What you are doing here is describing the x’ coordinate of an inertial system S’ of something moving at c in terms of the t coordinate in the inertial system S. It is unclear why you would want to do this and it certainly tells you nothing about an observer moving at c. The only inertial frames you have here are moving at a relative speed of v.

 

[PeterDonis]

You mean you cannot simplify before taking the limit v->c ?

As I was reading your OP, setting $x = ct$ is not taking the limit $v \rightarrow c$. It is setting $v = c$. I was assuming that the primed frame you were talking about was supposed to be a "rest frame" for the light ray you were describing with $x = ct$. With that interpretation, you are trying to cancel out factors that are zero, which is not well defined.

As @Orodruin and @PAllen are reading your OP, you are describing a light ray in the unprimed frame as $x = ct$, then transforming to a primed frame, then taking the limit $v \rightarrow c$ for the relative velocities of the two frames. With that interpretation, as @PAllen said, your expression is well-defined but you end up deducing that the entire light ray is a single event, which is incorrect.

So you are making a mistake either way, but which mistake you are making depends on how your ambiguous description in the OP is interpreted.

 

[jk22]

Thanks for the clarification.

 

[jk22]

But I still have a problem with the argument of not defined since the principle of the absolute invariant speed, the moving observer should have both speed 0 and c. How can a point have a nonzero speed relatively to itself, it would be 2 points then ?

Or should there a quantum delocalization argument be used ?

 

[Ibix]

How can a point have a nonzero speed relatively to itself,

It can't. Therefore an inertial frame moving at $c$ is a direct self-contradiction.

 

[PeterDonis]

I still have a problem with the argument of not defined

The Lorentz transformation not being well-defined for $v = c$ is not an "argument", it's a mathematical fact. If you have a problem with it, you need to retrain your intuitions; you can't "have a problem" with mathematical facts, at least not if you want to understand mathematical physics.

 

[robphy]

I still have a problem with the argument of not defined

The Lorentz transformation not being well-defined for $v = c$ is not an "argument", it's a mathematical fact. If you have a problem with it, you need to retrain your intuitions; you can't "have a problem" with mathematical facts, at least not if you want to understand mathematical physics.

The fundamental parameter in the Lorentz Transformation is not the velocity,
but it is
really the rapidity $\theta$ (an additive parameter with range $-\infty < \theta < \infty$ and where velocity $v=c\tanh\theta$ [which is not additive]).
The range $-\infty < \theta < \infty$ means that they are only associated with [necessarily, timelike] 4-velocities.

(Recall the boosts form a group... so every boost must have an inverse-boost.
No boost can get from some "rest" frame with $\theta=0$ to another with $\theta=\infty$
since there is no inverse boost from $\theta=\infty$ to $\theta=0$. [Note that $(\infty - \infty)$ is not well defined.])

 

[Herman Trivilino]

But I still have a problem with the argument of not defined

Look at it this way. Start with the postulate that a beam of light will recede from you at speed $c$ no matter how fast you chase after it. Therefore, no matter how large your speed $v$ you can never catch it. Thus $v$ can never equal $c$.

So, of course, if you set $v$ equal to $c$ in the Lorentz transformation equations you will get nonsense.